MCQ
$\mathop {\lim }\limits_{x \to 0} \,\frac{{{\rm{ln}}\,(\cos x)}}{{{x^2}}} = . . .$
- A$0$
- B$1$
- C$\frac{1}{2}$
- ✓$ - \frac{1}{2}$
$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}}$$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \tan x}}{{2x}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - {{\sec }^2}x}}{2} = \frac{{ - 1}}{2}$.
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